hypothesis of sample means

Teach yourself statistics

Hypothesis Test for a Mean

This lesson explains how to conduct a hypothesis test of a mean, when the following conditions are met:

• The sampling method is simple random sampling .
• The sampling distribution is normal or nearly normal.

Generally, the sampling distribution will be approximately normally distributed if any of the following conditions apply.

• The population distribution is normal.
• The population distribution is symmetric , unimodal , without outliers , and the sample size is 15 or less.
• The population distribution is moderately skewed , unimodal, without outliers, and the sample size is between 16 and 40.
• The sample size is greater than 40, without outliers.

This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

State the Hypotheses

Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis . The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false; and vice versa.

The table below shows three sets of hypotheses. Each makes a statement about how the population mean μ is related to a specified value M . (In the table, the symbol ≠ means " not equal to ".)

The first set of hypotheses (Set 1) is an example of a two-tailed test , since an extreme value on either side of the sampling distribution would cause a researcher to reject the null hypothesis. The other two sets of hypotheses (Sets 2 and 3) are one-tailed tests , since an extreme value on only one side of the sampling distribution would cause a researcher to reject the null hypothesis.

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

• Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
• Test method. Use the one-sample t-test to determine whether the hypothesized mean differs significantly from the observed sample mean.

Analyze Sample Data

Using sample data, conduct a one-sample t-test. This involves finding the standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic.

SE = s * sqrt{ ( 1/n ) * [ ( N - n ) / ( N - 1 ) ] }

SE = s / sqrt( n )

• Degrees of freedom. The degrees of freedom (DF) is equal to the sample size (n) minus one. Thus, DF = n - 1.

t = ( x - μ) / SE

• P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a t statistic, use the t Distribution Calculator to assess the probability associated with the t statistic, given the degrees of freedom computed above. (See sample problems at the end of this lesson for examples of how this is done.)

Sample Size Calculator

As you probably noticed, the process of hypothesis testing can be complex. When you need to test a hypothesis about a mean score, consider using the Sample Size Calculator. The calculator is fairly easy to use, and it is free. You can find the Sample Size Calculator in Stat Trek's main menu under the Stat Tools tab. Or you can tap the button below.

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level , and rejecting the null hypothesis when the P-value is less than the significance level.

Test Your Understanding

In this section, two sample problems illustrate how to conduct a hypothesis test of a mean score. The first problem involves a two-tailed test; the second problem, a one-tailed test.

Problem 1: Two-Tailed Test

An inventor has developed a new, energy-efficient lawn mower engine. He claims that the engine will run continuously for 5 hours (300 minutes) on a single gallon of regular gasoline. From his stock of 2000 engines, the inventor selects a simple random sample of 50 engines for testing. The engines run for an average of 295 minutes, with a standard deviation of 20 minutes. Test the null hypothesis that the mean run time is 300 minutes against the alternative hypothesis that the mean run time is not 300 minutes. Use a 0.05 level of significance. (Assume that run times for the population of engines are normally distributed.)

Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

Null hypothesis: μ = 300

Alternative hypothesis: μ ≠ 300

• Formulate an analysis plan . For this analysis, the significance level is 0.05. The test method is a one-sample t-test .

SE = s / sqrt(n) = 20 / sqrt(50) = 20/7.07 = 2.83

DF = n - 1 = 50 - 1 = 49

t = ( x - μ) / SE = (295 - 300)/2.83 = -1.77

where s is the standard deviation of the sample, x is the sample mean, μ is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test , the P-value is the probability that the t statistic having 49 degrees of freedom is less than -1.77 or greater than 1.77. We use the t Distribution Calculator to find P(t < -1.77) is about 0.04.

• If you enter 1.77 as the sample mean in the t Distribution Calculator, you will find the that the P(t < 1.77) is about 0.04. Therefore, P(t >  1.77) is 1 minus 0.96 or 0.04. Thus, the P-value = 0.04 + 0.04 = 0.08.
• Interpret results . Since the P-value (0.08) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the population was normally distributed, and the sample size was small relative to the population size (less than 5%).

Problem 2: One-Tailed Test

Bon Air Elementary School has 1000 students. The principal of the school thinks that the average IQ of students at Bon Air is at least 110. To prove her point, she administers an IQ test to 20 randomly selected students. Among the sampled students, the average IQ is 108 with a standard deviation of 10. Based on these results, should the principal accept or reject her original hypothesis? Assume a significance level of 0.01. (Assume that test scores in the population of engines are normally distributed.)

Null hypothesis: μ >= 110

Alternative hypothesis: μ < 110

• Formulate an analysis plan . For this analysis, the significance level is 0.01. The test method is a one-sample t-test .

SE = s / sqrt(n) = 10 / sqrt(20) = 10/4.472 = 2.236

DF = n - 1 = 20 - 1 = 19

t = ( x - μ) / SE = (108 - 110)/2.236 = -0.894

Here is the logic of the analysis: Given the alternative hypothesis (μ < 110), we want to know whether the observed sample mean is small enough to cause us to reject the null hypothesis.

The observed sample mean produced a t statistic test statistic of -0.894. We use the t Distribution Calculator to find P(t < -0.894) is about 0.19.

• This means we would expect to find a sample mean of 108 or smaller in 19 percent of our samples, if the true population IQ were 110. Thus the P-value in this analysis is 0.19.
• Interpret results . Since the P-value (0.19) is greater than the significance level (0.01), we cannot reject the null hypothesis.

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Hypothesis Testing | A Step-by-Step Guide with Easy Examples

Published on November 8, 2019 by Rebecca Bevans . Revised on June 22, 2023.

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics . It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.

There are 5 main steps in hypothesis testing:

• State your research hypothesis as a null hypothesis and alternate hypothesis (H o ) and (H a  or H 1 ).
• Collect data in a way designed to test the hypothesis.
• Perform an appropriate statistical test .
• Decide whether to reject or fail to reject your null hypothesis.
• Present the findings in your results and discussion section.

Though the specific details might vary, the procedure you will use when testing a hypothesis will always follow some version of these steps.

Table of contents

Step 1: state your null and alternate hypothesis, step 2: collect data, step 3: perform a statistical test, step 4: decide whether to reject or fail to reject your null hypothesis, step 5: present your findings, other interesting articles, frequently asked questions about hypothesis testing.

After developing your initial research hypothesis (the prediction that you want to investigate), it is important to restate it as a null (H o ) and alternate (H a ) hypothesis so that you can test it mathematically.

The alternate hypothesis is usually your initial hypothesis that predicts a relationship between variables. The null hypothesis is a prediction of no relationship between the variables you are interested in.

• H 0 : Men are, on average, not taller than women. H a : Men are, on average, taller than women.

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For a statistical test to be valid , it is important to perform sampling and collect data in a way that is designed to test your hypothesis. If your data are not representative, then you cannot make statistical inferences about the population you are interested in.

There are a variety of statistical tests available, but they are all based on the comparison of within-group variance (how spread out the data is within a category) versus between-group variance (how different the categories are from one another).

If the between-group variance is large enough that there is little or no overlap between groups, then your statistical test will reflect that by showing a low p -value . This means it is unlikely that the differences between these groups came about by chance.

Alternatively, if there is high within-group variance and low between-group variance, then your statistical test will reflect that with a high p -value. This means it is likely that any difference you measure between groups is due to chance.

Your choice of statistical test will be based on the type of variables and the level of measurement of your collected data .

• an estimate of the difference in average height between the two groups.
• a p -value showing how likely you are to see this difference if the null hypothesis of no difference is true.

Based on the outcome of your statistical test, you will have to decide whether to reject or fail to reject your null hypothesis.

In most cases you will use the p -value generated by your statistical test to guide your decision. And in most cases, your predetermined level of significance for rejecting the null hypothesis will be 0.05 – that is, when there is a less than 5% chance that you would see these results if the null hypothesis were true.

In some cases, researchers choose a more conservative level of significance, such as 0.01 (1%). This minimizes the risk of incorrectly rejecting the null hypothesis ( Type I error ).

The results of hypothesis testing will be presented in the results and discussion sections of your research paper , dissertation or thesis .

In the results section you should give a brief summary of the data and a summary of the results of your statistical test (for example, the estimated difference between group means and associated p -value). In the discussion , you can discuss whether your initial hypothesis was supported by your results or not.

In the formal language of hypothesis testing, we talk about rejecting or failing to reject the null hypothesis. You will probably be asked to do this in your statistics assignments.

However, when presenting research results in academic papers we rarely talk this way. Instead, we go back to our alternate hypothesis (in this case, the hypothesis that men are on average taller than women) and state whether the result of our test did or did not support the alternate hypothesis.

If your null hypothesis was rejected, this result is interpreted as “supported the alternate hypothesis.”

These are superficial differences; you can see that they mean the same thing.

You might notice that we don’t say that we reject or fail to reject the alternate hypothesis . This is because hypothesis testing is not designed to prove or disprove anything. It is only designed to test whether a pattern we measure could have arisen spuriously, or by chance.

If we reject the null hypothesis based on our research (i.e., we find that it is unlikely that the pattern arose by chance), then we can say our test lends support to our hypothesis . But if the pattern does not pass our decision rule, meaning that it could have arisen by chance, then we say the test is inconsistent with our hypothesis .

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Normal distribution
• Descriptive statistics
• Measures of central tendency
• Correlation coefficient

Methodology

• Cluster sampling
• Stratified sampling
• Types of interviews
• Cohort study
• Thematic analysis

Research bias

• Implicit bias
• Cognitive bias
• Survivorship bias
• Availability heuristic
• Nonresponse bias
• Regression to the mean

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance.

A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question.

A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

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8.4.3 Hypothesis Testing for the Mean

$\quad$ $H_0$: $\mu=\mu_0$, $\quad$ $H_1$: $\mu \neq \mu_0$.

$\quad$ $H_0$: $\mu \leq \mu_0$, $\quad$ $H_1$: $\mu > \mu_0$.

$\quad$ $H_0$: $\mu \geq \mu_0$, $\quad$ $H_1$: $\mu \lt \mu_0$.

Two-sided Tests for the Mean:

Therefore, we can suggest the following test. Choose a threshold, and call it $c$. If $|W| \leq c$, accept $H_0$, and if $|W|>c$, accept $H_1$. How do we choose $c$? If $\alpha$ is the required significance level, we must have

• As discussed above, we let \begin{align}%\label{} W(X_1,X_2, \cdots,X_n)=\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}. \end{align} Note that, assuming $H_0$, $W \sim N(0,1)$. We will choose a threshold, $c$. If $|W| \leq c$, we accept $H_0$, and if $|W|>c$, accept $H_1$. To choose $c$, we let \begin{align} P(|W| > c \; | \; H_0) =\alpha. \end{align} Since the standard normal PDF is symmetric around $0$, we have \begin{align} P(|W| > c \; | \; H_0) = 2 P(W>c | \; H_0). \end{align} Thus, we conclude $P(W>c | \; H_0)=\frac{\alpha}{2}$. Therefore, \begin{align} c=z_{\frac{\alpha}{2}}. \end{align} Therefore, we accept $H_0$ if \begin{align} \left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \leq z_{\frac{\alpha}{2}}, \end{align} and reject it otherwise.
• We have \begin{align} \beta (\mu) &=P(\textrm{type II error}) = P(\textrm{accept }H_0 \; | \; \mu) \\ &= P\left(\left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \lt z_{\frac{\alpha}{2}}\; | \; \mu \right). \end{align} If $X_i \sim N(\mu,\sigma^2)$, then $\overline{X} \sim N(\mu, \frac{\sigma^2}{n})$. Thus, \begin{align} \beta (\mu)&=P\left(\left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \lt z_{\frac{\alpha}{2}}\; | \; \mu \right)\\ &=P\left(\mu_0- z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \leq \overline{X} \leq \mu_0+ z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}\right)\\ &=\Phi\left(z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right)-\Phi\left(-z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right). \end{align}
• Let $S^2$ be the sample variance for this random sample. Then, the random variable $W$ defined as \begin{equation} W(X_1,X_2, \cdots, X_n)=\frac{\overline{X}-\mu_0}{S / \sqrt{n}} \end{equation} has a $t$-distribution with $n-1$ degrees of freedom, i.e., $W \sim T(n-1)$. Thus, we can repeat the analysis of Example 8.24 here. The only difference is that we need to replace $\sigma$ by $S$ and $z_{\frac{\alpha}{2}}$ by $t_{\frac{\alpha}{2},n-1}$. Therefore, we accept $H_0$ if \begin{align} |W| \leq t_{\frac{\alpha}{2},n-1}, \end{align} and reject it otherwise. Let us look at a numerical example of this case.

$\quad$ $H_0$: $\mu=170$, $\quad$ $H_1$: $\mu \neq 170$.

• Let's first compute the sample mean and the sample standard deviation. The sample mean is \begin{align}%\label{} \overline{X}&=\frac{X_1+X_2+X_3+X_4+X_5+X_6+X_7+X_8+X_9}{9}\\ &=165.8 \end{align} The sample variance is given by \begin{align}%\label{} {S}^2=\frac{1}{9-1} \sum_{k=1}^9 (X_k-\overline{X})^2&=68.01 \end{align} The sample standard deviation is given by \begin{align}%\label{} S&= \sqrt{S^2}=8.25 \end{align} The following MATLAB code can be used to obtain these values: x=[176.2,157.9,160.1,180.9,165.1,167.2,162.9,155.7,166.2]; m=mean(x); v=var(x); s=std(x); Now, our test statistic is \begin{align} W(X_1,X_2, \cdots, X_9)&=\frac{\overline{X}-\mu_0}{S / \sqrt{n}}\\ &=\frac{165.8-170}{8.25 / 3}=-1.52 \end{align} Thus, $|W|=1.52$. Also, we have \begin{align} t_{\frac{\alpha}{2},n-1} = t_{0.025,8} \approx 2.31 \end{align} The above value can be obtained in MATLAB using the command $\mathtt{tinv(0.975,8)}$. Thus, we conclude \begin{align} |W| \leq t_{\frac{\alpha}{2},n-1}. \end{align} Therefore, we accept $H_0$. In other words, we do not have enough evidence to conclude that the average height in the city is different from the average height in the country.

Let us summarize what we have obtained for the two-sided test for the mean.

One-sided Tests for the Mean:

• As before, we define the test statistic as \begin{align}%\label{} W(X_1,X_2, \cdots,X_n)=\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}. \end{align} If $H_0$ is true (i.e., $\mu \leq \mu_0$), we expect $\overline{X}$ (and thus $W$) to be relatively small, while if $H_1$ is true, we expect $\overline{X}$ (and thus $W$) to be larger. This suggests the following test: Choose a threshold, and call it $c$. If $W \leq c$, accept $H_0$, and if $W>c$, accept $H_1$. How do we choose $c$? If $\alpha$ is the required significance level, we must have \begin{align} P(\textrm{type I error}) &= P(\textrm{Reject }H_0 \; | \; H_0) \\ &= P(W > c \; | \; \mu \leq \mu_0) \leq \alpha. \end{align} Here, the probability of type I error depends on $\mu$. More specifically, for any $\mu \leq \mu_0$, we can write \begin{align} P(\textrm{type I error} \; | \; \mu) &= P(\textrm{Reject }H_0 \; | \; \mu) \\ &= P(W > c \; | \; \mu)\\ &=P \left(\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}> c \; | \; \mu\right)\\ &=P \left(\frac{\overline{X}-\mu}{\sigma / \sqrt{n}}+\frac{\mu-\mu_0}{\sigma / \sqrt{n}}> c \; | \; \mu\right)\\ &=P \left(\frac{\overline{X}-\mu}{\sigma / \sqrt{n}}> c+\frac{\mu_0-\mu}{\sigma / \sqrt{n}} \; | \; \mu\right)\\ &\leq P \left(\frac{\overline{X}-\mu}{\sigma / \sqrt{n}}> c \; | \; \mu\right) \quad (\textrm{ since }\mu \leq \mu_0)\\ &=1-\Phi(c) \quad \big(\textrm{ since given }\mu, \frac{\overline{X}-\mu}{\sigma / \sqrt{n}} \sim N(0,1) \big). \end{align} Thus, we can choose $\alpha=1-\Phi(c)$, which results in \begin{align} c=z_{\alpha}. \end{align} Therefore, we accept $H_0$ if \begin{align} \frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \leq z_{\alpha}, \end{align} and reject it otherwise.

$\quad$ $H_0$: $\mu \geq \mu_0$, $\quad$ $H_1$: $\mu \lt \mu_0$,

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8.6 Hypothesis Tests for a Population Mean with Known Population Standard Deviation

Learning objectives.

• Conduct and interpret hypothesis tests for a population mean with known population standard deviation.

Some notes about conducting a hypothesis test:

• The null hypothesis [latex]H_0[/latex] is always an “equal to.”  The null hypothesis is the original claim about the population parameter.
• The alternative hypothesis [latex]H_a[/latex] is a “less than,” “greater than,” or “not equal to.”  The form of the alternative hypothesis depends on the context of the question.
• If the alternative hypothesis is a “less than”, then the test is left-tail.  The p -value is the area in the left-tail of the distribution.
• If the alternative hypothesis is a “greater than”, then the test is right-tail.  The p -value is the area in the right-tail of the distribution.
• If the alternative hypothesis is a “not equal to”, then the test is two-tail.  The p -value is the sum of the area in the two-tails of the distribution.  Each tail represents exactly half of the p -value.
• Think about the meaning of the p -value.  A data analyst (and anyone else) should have more confidence that they made the correct decision to reject the null hypothesis with a smaller p -value (for example, 0.001 as opposed to 0.04) even if using a significance level of 0.05.  Similarly, for a large p -value such as 0.4, as opposed to a p -value of 0.056 (a significance level of 0.05 is less than either number), a data analyst should have more confidence that they made the correct decision in not rejecting the null hypothesis.  This makes the data analyst use judgment rather than mindlessly applying rules.
• The significance level must be identified before collecting the sample data and conducting the test.  Generally, the significance level will be included in the question.  If no significance level is given, a common standard is to use a significance level of 5%.
• An alternative approach for hypothesis testing is to use what is called the critical value approach .  In this book, we will only use the p -value approach.  Some of the videos below may mention the critical value approach, but this approach will not be used in this book.

Suppose the hypotheses for a hypothesis test are:

[latex]\begin{eqnarray*} H_0: & & \mu=5 \\ H_a: & & \mu \lt 5 \end{eqnarray*}[/latex]

Because the alternative hypothesis is a [latex]\lt[/latex], this is a left-tailed test.  The p -value is the area in the left-tail of the distribution.

[latex]\begin{eqnarray*} H_0: & & \mu=0.5 \\ H_a: & & \mu \neq 0.5  \end{eqnarray*}[/latex]

Because the alternative hypothesis is a [latex]\neq[/latex], this is a two-tailed test.  The p -value is the sum of the areas in the two tails of the distribution.  Each tail contains exactly half of the p -value.

[latex]\begin{eqnarray*} H_0: & & \mu=10 \\ H_a: & & \mu \lt 10  \end{eqnarray*}[/latex]

Steps to Conduct a Hypothesis Test for a Population Mean with Known Population Standard Deviation

• Write down the null and alternative hypotheses in terms of the population mean [latex]\mu[/latex].  Include appropriate units with the values of the mean.
• Use the form of the alternative hypothesis to determine if the test is left-tailed, right-tailed, or two-tailed.
• Collect the sample information for the test and identify the significance level [latex]\alpha[/latex].
• When the population standard deviation is known , we use a normal distribution with [latex]\displaystyle{z=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}}[/latex] to find the p -value.  The p -value is the area in the corresponding tail of the normal distribution.
• The results of the sample data are significant.  There is sufficient evidence to conclude that the null hypothesis [latex]H_0[/latex] is an incorrect belief and that the alternative hypothesis [latex]H_a[/latex] is most likely correct.
• The results of the sample data are not significant.  There is not sufficient evidence to conclude that the alternative hypothesis [latex]H_a[/latex] may be correct.
• Write down a concluding sentence specific to the context of the question.

USING EXCEL TO CALCULE THE P -VALUE FOR A HYPOTHESIS TEST ON A POPULATION MEAN WITH KNOWN POPULATION STANDARD DEVIATION

The p -value for a hypothesis test on a population mean is the area in the tail(s) of the distribution of the sample mean.  When the population standard deviation is known, use the normal distribution to find the p -value.

The p -value is the area in the tail(s) of a normal distribution, so the norm.dist(x,[latex]\mu[/latex],[latex]\sigma[/latex],logic operator) function can be used to calculate the p -value.

• For x , enter the value for [latex]\overline{x}[/latex].
• For [latex]\mu[/latex] , enter the mean of the sample means [latex]\mu[/latex].  Note:  Because the test is run assuming the null hypothesis is true, the value for [latex]\mu[/latex] is the claim from the null hypothesis.
• For [latex]\sigma[/latex] , enter the standard error of the mean [latex]\displaystyle{\frac{\sigma}{\sqrt{n}}}[/latex].
• For the logic operator , enter true .  Note:  Because we are calculating the area under the curve, we always enter true for the logic operator.

Use the appropriate technique with the norm.dist function to find the area in the left-tail or the area in the right-tail.

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds with a standard deviation of 0.8 seconds for swimming the 25-meter freestyle.  His dad, Frank, thought that Jeffrey could swim the 25-meter freestyle faster using goggles.  Frank bought Jeffrey a new pair of goggles and timed Jeffrey swimming the 25-meter freestyle 15 different times.  In the sample of 15 swims, Jeffrey’s mean time was 16 seconds.  Frank thought that the goggles helped Jeffrey swim faster than 16.43 seconds.  At the 5% significance level, did Jeffrey swim faster wearing the goggles?  Assume that the swim times for the 25-meter freestyle are normally distributed.

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu=16.43 \mbox{ seconds} \\ H_a: & & \mu \lt 16.43 \mbox{ seconds} \end{eqnarray*}[/latex]

From the question, we have [latex]n=15[/latex], [latex]\overline{x}=16[/latex], [latex]\sigma=0.8[/latex] and [latex]\alpha=0.05[/latex].

This is a test on a population mean where the population standard deviation is known ([latex]\sigma=0.8[/latex]).  So we use a normal distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\lt[/latex], the p -value is the area in the left-tail of the distribution.

So the p -value[latex]=0.0187[/latex].

Conclusion:

Because p -value[latex]=0.0187 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that Jeffrey’s mean swim time with the goggles is less than 16.43 seconds.

• The null hypothesis [latex]\mu=16.43[/latex] is the claim that Jeffrey’s mean swim time with the goggles is 16.43 seconds (the same as it is without the googles).
• The alternative hypothesis [latex]\mu \lt 16.43[/latex] is the claim that Jeffrey’s swim time with the goggles is less than 16.43 seconds.
• The function is norm.dist because we are finding the area in the left tail of a normal distribution.
• Field 1 is the value of [latex]\overline{x}[/latex]
• Field 2 is the value of [latex]\mu[/latex] from the null hypothesis.  Remember, we run the test assuming the null hypothesis is true, so that means we assume [latex]\mu=16.43[/latex].
• Field 3 is the standard deviation for the sample means [latex]\displaystyle{\frac{\sigma}{\sqrt{n}}}[/latex].  Note that we are not using the standard deviation from the population ([latex]\sigma=0.8[/latex]).  This is because the p -value is the area under the curve of the distribution of the sample means, not the distribution of the population.
• The p -value of 0.0187 tells us that under the assumption that Jeffrey’s mean swim time with goggles is 16.43 seconds (the null hypothesis), there is only a 1.87% chance that the mean time for the 15 sample swims is 16 seconds or less.  This is a small probability, and so is unlikely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely incorrect, and so the conclusion of the test is to reject the null hypothesis in favour of the alternative hypothesis.
• The Type I error for this problem is to conclude that Jeffrey swims the 25-meter freestyle, on average, in less than 16.43 seconds (the alternative hypothesis) when, in fact, he actually swims the 25-meter freestyle, on average, in 16.43 seconds (the null hypothesis).  That is, reject the null hypothesis when the null hypothesis is actually true.
• The Type II error for this problem is to conclude that Jeffrey swims the 25-meter freestyle, on average, in 16.43 seconds (the null hypothesis) when, in fact, he actually swims the 25-meter freestyle, on average, in less than 16.43 seconds (the alternative hypothesis).  That is, do not reject the null hypothesis when the null hypothesis is actually false.

The mean throwing distance of a football for Marco, a high school freshman quarterback, is 40 yards with a standard deviation of 2 yards.  The team coach tells Marco to adjust his grip to get more distance.  The coach records the distances for 20 throws with the new grip.  For the 20 throws, Marco’s mean distance was 41.5 yards.  The coach thought the different grip helped Marco throw farther than 40 yards.  At the 5% significance level, is Marco’s mean throwing distance higher with the new grip?  Assume the throw distances for footballs are normally distributed.

[latex]\begin{eqnarray*} H_0: & & \mu=40 \mbox{ yards} \\ H_a: & & \mu \gt 40 \mbox{ yards} \end{eqnarray*}[/latex]

From the question, we have [latex]n=20[/latex], [latex]\overline{x}=41.5[/latex], [latex]\sigma=2[/latex] and [latex]\alpha=0.05[/latex].

This is a test on a population mean where the population standard deviation is known ([latex]\sigma=2[/latex]).  So we use a normal distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\gt[/latex], the p -value is the area in the right-tail of the distribution.

So the p -value[latex]=0.0004[/latex].

Because p -value[latex]=0.0004 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that Marco’s mean throwing distance is greater than 40 yards with the new grip.

• The null hypothesis [latex]\mu=40[/latex] is the claim that Marco’s mean throwing distance with the new grip is 40 yards (the same as it is without the new grip).
• The alternative hypothesis [latex]\mu \gt 40[/latex] is the claim that Marco’s mean throwing distance with the new grip is greater than 40 yards.
• Field 2 is the value of [latex]\mu[/latex] from the null hypothesis.
• Field 3 is the standard deviation for the sample means [latex]\displaystyle{\frac{\sigma}{\sqrt{n}}}[/latex].
• The p -value of 0.0004 tells us that under the assumption that Marco’s mean throwing distance with the new grip is 40 yards, there is only a 0.047% chance that the mean throwing distance for the 20 sample throws is more than 40 yards.  This is a small probability, and so is unlikely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely incorrect, and so the conclusion of the test is to reject the null hypothesis in favour of the alternative hypothesis.

A local college states in its marketing materials that the average age of its first-year students is 18.3 years with a standard deviation of 3.4 years.  But this information is based on old data and does not take into account that more older adults are returning to college.  A researcher at the college believes that the average age of its first-year students has changed.  The researcher takes a sample of 50 first-year students and finds the average age is 19.5 years.  At the 1% significance level, has the average age of the college’s first-year students changed?

[latex]\begin{eqnarray*} H_0: & & \mu=18.3 \mbox{ years} \\ H_a: & & \mu \neq 18.3 \mbox{ years} \end{eqnarray*}[/latex]

From the question, we have [latex]n=50[/latex], [latex]\overline{x}=19.5[/latex], [latex]\sigma=3.4[/latex] and [latex]\alpha=0.01[/latex].

This is a test on a population mean where the population standard deviation is known ([latex]\sigma=3.4[/latex]).  In this case, the sample size is greater than 30.  So we use a normal distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\neq[/latex], the p -value is the sum of area in the tails of the distribution.

Because there is only one sample, we only have information relating to one of the two tails, either the left tail or the right tail.  We need to know if the sample relates to the left tail or right tail because that will determine how we calculate out the area of that tail using the normal distribution.  In this case, the sample mean [latex]\overline{x}=19.5[/latex] is greater than the value of the population mean in the null hypothesis [latex]\mu=18.3[/latex] ([latex]\overline{x}=19.5>18.3=\mu[/latex]), so the sample information relates to the right-tail of the normal distribution.  This means that we will calculate out the area in the right tail using 1-norm.dist .  However, this is a two-tailed test where the p -value is the sum of the area in the two tails and the area in the right-tail is only one half of the p -value.  The area in the left tail equals the area in the right tail and the p -value is the sum of these two areas.

So the area in the right tail is 0.0063 and [latex]\frac{1}{2}[/latex]( p -value)[latex]=0.0063[/latex].  This is also the area in the left tail, so

p -value[latex]=0.0063+0.0063=0.0126[/latex]

Because p -value[latex]=0.0126 \gt 0.01=\alpha[/latex], we do not reject the null hypothesis.  At the 1% significance level there is not enough evidence to suggest that the average age of the college’s first-year students has changed.

• The null hypothesis [latex]\mu=18.3[/latex] is the claim that the average age of the first-year students is still 18.3 years.
• The alternative hypothesis [latex]\mu \neq 18.3[/latex] is the claim that the average age of the first-year students has changed from 18.3 years.
• We use norm.dist([latex]\overline{x}[/latex],[latex]\mu[/latex],[latex]\sigma/\mbox{sqrt}(n)[/latex],true) to find the area in the left tail.  The area in the right tail equals the area in the left tail, so we can find the p -value by adding the output from this function to itself.
• We use 1-norm.dist([latex]\overline{x}[/latex],[latex]\mu[/latex],[latex]\sigma/\mbox{sqrt}(n)[/latex],true) to find the area in the right tail.  The area in the left tail equals the area in the right tail, so we can find the p -value by adding the output from this function to itself.
• The p -value of 0.0126  is a large probability compared to the 1% significance level, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.  In other words, the claim that the average age of first-year students is 18.3 years is most likely correct.

Watch this video: Hypothesis Testing: z -test, right tail by ExcelIsFun [33:47]

Watch this video: Hypothesis Testing: z -test, left tail by ExcelIsFun [10:57]

Watch this video: Hypothesis Testing: z -test, two tail by ExcelIsFun [9:56]

Concept Review

The hypothesis test for a population mean is a well established process:

• Collect the sample information for the test and identify the significance level.
• When the population standard deviation is known, find the p -value (the area in the corresponding tail) for the test using the normal distribution.
• Compare the p -value to the significance level and state the outcome of the test.

Attribution

“ 9.6   Hypothesis Testing of a Single Mean and Single Proportion “ in Introductory Statistics by OpenStax  is licensed under a  Creative Commons Attribution 4.0 International License.

Introduction to Statistics Copyright © 2022 by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Statistics By Jim

Making statistics intuitive

Statistical Hypothesis Testing Overview

By Jim Frost 59 Comments

In this blog post, I explain why you need to use statistical hypothesis testing and help you navigate the essential terminology. Hypothesis testing is a crucial procedure to perform when you want to make inferences about a population using a random sample. These inferences include estimating population properties such as the mean, differences between means, proportions, and the relationships between variables.

This post provides an overview of statistical hypothesis testing. If you need to perform hypothesis tests, consider getting my book, Hypothesis Testing: An Intuitive Guide .

Why You Should Perform Statistical Hypothesis Testing

Hypothesis testing is a form of inferential statistics that allows us to draw conclusions about an entire population based on a representative sample. You gain tremendous benefits by working with a sample. In most cases, it is simply impossible to observe the entire population to understand its properties. The only alternative is to collect a random sample and then use statistics to analyze it.

While samples are much more practical and less expensive to work with, there are trade-offs. When you estimate the properties of a population from a sample, the sample statistics are unlikely to equal the actual population value exactly.  For instance, your sample mean is unlikely to equal the population mean. The difference between the sample statistic and the population value is the sample error.

Differences that researchers observe in samples might be due to sampling error rather than representing a true effect at the population level. If sampling error causes the observed difference, the next time someone performs the same experiment the results might be different. Hypothesis testing incorporates estimates of the sampling error to help you make the correct decision. Learn more about Sampling Error .

For example, if you are studying the proportion of defects produced by two manufacturing methods, any difference you observe between the two sample proportions might be sample error rather than a true difference. If the difference does not exist at the population level, you won’t obtain the benefits that you expect based on the sample statistics. That can be a costly mistake!

Let’s cover some basic hypothesis testing terms that you need to know.

Background information : Difference between Descriptive and Inferential Statistics and Populations, Parameters, and Samples in Inferential Statistics

Hypothesis Testing

Hypothesis testing is a statistical analysis that uses sample data to assess two mutually exclusive theories about the properties of a population. Statisticians call these theories the null hypothesis and the alternative hypothesis. A hypothesis test assesses your sample statistic and factors in an estimate of the sample error to determine which hypothesis the data support.

When you can reject the null hypothesis, the results are statistically significant, and your data support the theory that an effect exists at the population level.

The effect is the difference between the population value and the null hypothesis value. The effect is also known as population effect or the difference. For example, the mean difference between the health outcome for a treatment group and a control group is the effect.

Typically, you do not know the size of the actual effect. However, you can use a hypothesis test to help you determine whether an effect exists and to estimate its size. Hypothesis tests convert your sample effect into a test statistic, which it evaluates for statistical significance. Learn more about Test Statistics .

An effect can be statistically significant, but that doesn’t necessarily indicate that it is important in a real-world, practical sense. For more information, read my post about Statistical vs. Practical Significance .

Null Hypothesis

The null hypothesis is one of two mutually exclusive theories about the properties of the population in hypothesis testing. Typically, the null hypothesis states that there is no effect (i.e., the effect size equals zero). The null is often signified by H 0 .

In all hypothesis testing, the researchers are testing an effect of some sort. The effect can be the effectiveness of a new vaccination, the durability of a new product, the proportion of defect in a manufacturing process, and so on. There is some benefit or difference that the researchers hope to identify.

However, it’s possible that there is no effect or no difference between the experimental groups. In statistics, we call this lack of an effect the null hypothesis. Therefore, if you can reject the null, you can favor the alternative hypothesis, which states that the effect exists (doesn’t equal zero) at the population level.

You can think of the null as the default theory that requires sufficiently strong evidence against in order to reject it.

For example, in a 2-sample t-test, the null often states that the difference between the two means equals zero.

When you can reject the null hypothesis, your results are statistically significant. Learn more about Statistical Significance: Definition & Meaning .

Related post : Understanding the Null Hypothesis in More Detail

Alternative Hypothesis

The alternative hypothesis is the other theory about the properties of the population in hypothesis testing. Typically, the alternative hypothesis states that a population parameter does not equal the null hypothesis value. In other words, there is a non-zero effect. If your sample contains sufficient evidence, you can reject the null and favor the alternative hypothesis. The alternative is often identified with H 1 or H A .

For example, in a 2-sample t-test, the alternative often states that the difference between the two means does not equal zero.

You can specify either a one- or two-tailed alternative hypothesis:

If you perform a two-tailed hypothesis test, the alternative states that the population parameter does not equal the null value. For example, when the alternative hypothesis is H A : μ ≠ 0, the test can detect differences both greater than and less than the null value.

A one-tailed alternative has more power to detect an effect but it can test for a difference in only one direction. For example, H A : μ > 0 can only test for differences that are greater than zero.

Related posts : Understanding T-tests and One-Tailed and Two-Tailed Hypothesis Tests Explained

P-values are the probability that you would obtain the effect observed in your sample, or larger, if the null hypothesis is correct. In simpler terms, p-values tell you how strongly your sample data contradict the null. Lower p-values represent stronger evidence against the null. You use P-values in conjunction with the significance level to determine whether your data favor the null or alternative hypothesis.

Related post : Interpreting P-values Correctly

Significance Level (Alpha)

For instance, a significance level of 0.05 signifies a 5% risk of deciding that an effect exists when it does not exist.

Use p-values and significance levels together to help you determine which hypothesis the data support. If the p-value is less than your significance level, you can reject the null and conclude that the effect is statistically significant. In other words, the evidence in your sample is strong enough to be able to reject the null hypothesis at the population level.

Related posts : Graphical Approach to Significance Levels and P-values and Conceptual Approach to Understanding Significance Levels

Types of Errors in Hypothesis Testing

Statistical hypothesis tests are not 100% accurate because they use a random sample to draw conclusions about entire populations. There are two types of errors related to drawing an incorrect conclusion.

• False positives: You reject a null that is true. Statisticians call this a Type I error . The Type I error rate equals your significance level or alpha (α).
• False negatives: You fail to reject a null that is false. Statisticians call this a Type II error. Generally, you do not know the Type II error rate. However, it is a larger risk when you have a small sample size , noisy data, or a small effect size. The type II error rate is also known as beta (β).

Statistical power is the probability that a hypothesis test correctly infers that a sample effect exists in the population. In other words, the test correctly rejects a false null hypothesis. Consequently, power is inversely related to a Type II error. Power = 1 – β. Learn more about Power in Statistics .

Related posts : Types of Errors in Hypothesis Testing and Estimating a Good Sample Size for Your Study Using Power Analysis

Which Type of Hypothesis Test is Right for You?

There are many different types of procedures you can use. The correct choice depends on your research goals and the data you collect. Do you need to understand the mean or the differences between means? Or, perhaps you need to assess proportions. You can even use hypothesis testing to determine whether the relationships between variables are statistically significant.

To choose the proper statistical procedure, you’ll need to assess your study objectives and collect the correct type of data . This background research is necessary before you begin a study.

Related Post : Hypothesis Tests for Continuous, Binary, and Count Data

Statistical tests are crucial when you want to use sample data to make conclusions about a population because these tests account for sample error. Using significance levels and p-values to determine when to reject the null hypothesis improves the probability that you will draw the correct conclusion.

To see an alternative approach to these traditional hypothesis testing methods, learn about bootstrapping in statistics !

If you want to see examples of hypothesis testing in action, I recommend the following posts that I have written:

• How Effective Are Flu Shots? This example shows how you can use statistics to test proportions.
• Fatality Rates in Star Trek . This example shows how to use hypothesis testing with categorical data.
• Busting Myths About the Battle of the Sexes . A fun example based on a Mythbusters episode that assess continuous data using several different tests.
• Are Yawns Contagious? Another fun example inspired by a Mythbusters episode.

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Reader Interactions

January 14, 2024 at 8:43 am

Hello professor Jim, how are you doing! Pls. What are the properties of a population and their examples? Thanks for your time and understanding.

January 14, 2024 at 12:57 pm

Please read my post about Populations vs. Samples for more information and examples.

Also, please note there is a search bar in the upper-right margin of my website. Use that to search for topics.

July 5, 2023 at 7:05 am

Hello, I have a question as I read your post. You say in p-values section

“P-values are the probability that you would obtain the effect observed in your sample, or larger, if the null hypothesis is correct. In simpler terms, p-values tell you how strongly your sample data contradict the null. Lower p-values represent stronger evidence against the null.”

But according to your definition of effect, the null states that an effect does not exist, correct? So what I assume you want to say is that “P-values are the probability that you would obtain the effect observed in your sample, or larger, if the null hypothesis is **incorrect**.”

July 6, 2023 at 5:18 am

Hi Shrinivas,

The correct definition of p-value is that it is a probability that exists in the context of a true null hypothesis. So, the quotation is correct in stating “if the null hypothesis is correct.”

Essentially, the p-value tells you the likelihood of your observed results (or more extreme) if the null hypothesis is true. It gives you an idea of whether your results are surprising or unusual if there is no effect.

Hence, with sufficiently low p-values, you reject the null hypothesis because it’s telling you that your sample results were unlikely to have occurred if there was no effect in the population.

I hope that helps make it more clear. If not, let me know I’ll attempt to clarify!

May 8, 2023 at 12:47 am

Thanks a lot Ny best regards

May 7, 2023 at 11:15 pm

Hi Jim Can you tell me something about size effect? Thanks

May 8, 2023 at 12:29 am

Here’s a post that I’ve written about Effect Sizes that will hopefully tell you what you need to know. Please read that. Then, if you have any more specific questions about effect sizes, please post them there. Thanks!

January 7, 2023 at 4:19 pm

Hi Jim, I have only read two pages so far but I am really amazed because in few paragraphs you made me clearly understand the concepts of months of courses I received in biostatistics! Thanks so much for this work you have done it helps a lot!

January 10, 2023 at 3:25 pm

Thanks so much!

June 17, 2021 at 1:45 pm

Can you help in the following question: Rocinante36 is priced at ₹7 lakh and has been designed to deliver a mileage of 22 km/litre and a top speed of 140 km/hr. Formulate the null and alternative hypotheses for mileage and top speed to check whether the new models are performing as per the desired design specifications.

April 19, 2021 at 1:51 pm

Its indeed great to read your work statistics.

I have a doubt regarding the one sample t-test. So as per your book on hypothesis testing with reference to page no 45, you have mentioned the difference between “the sample mean and the hypothesised mean is statistically significant”. So as per my understanding it should be quoted like “the difference between the population mean and the hypothesised mean is statistically significant”. The catch here is the hypothesised mean represents the sample mean.

Please help me understand this.

Regards Rajat

April 19, 2021 at 3:46 pm

Thanks for buying my book. I’m so glad it’s been helpful!

The test is performed on the sample but the results apply to the population. Hence, if the difference between the sample mean (observed in your study) and the hypothesized mean is statistically significant, that suggests that population does not equal the hypothesized mean.

For one sample tests, the hypothesized mean is not the sample mean. It is a mean that you want to use for the test value. It usually represents a value that is important to your research. In other words, it’s a value that you pick for some theoretical/practical reasons. You pick it because you want to determine whether the population mean is different from that particular value.

I hope that helps!

November 5, 2020 at 6:24 am

Jim, you are such a magnificent statistician/economist/econometrician/data scientist etc whatever profession. Your work inspires and simplifies the lives of so many researchers around the world. I truly admire you and your work. I will buy a copy of each book you have on statistics or econometrics. Keep doing the good work. Remain ever blessed

November 6, 2020 at 9:47 pm

Hi Renatus,

Thanks so much for you very kind comments. You made my day!! I’m so glad that my website has been helpful. And, thanks so much for supporting my books! 🙂

November 2, 2020 at 9:32 pm

Hi Jim, I hope you are aware of 2019 American Statistical Association’s official statement on Statistical Significance: https://www.tandfonline.com/doi/full/10.1080/00031305.2019.1583913 In case you do not bother reading the full article, may I quote you the core message here: “We conclude, based on our review of the articles in this special issue and the broader literature, that it is time to stop using the term “statistically significant” entirely. Nor should variants such as “significantly different,” “p < 0.05,” and “nonsignificant” survive, whether expressed in words, by asterisks in a table, or in some other way."

With best wishes,

November 3, 2020 at 2:09 am

I’m definitely aware of the debate surrounding how to use p-values most effectively. However, I need to correct you on one point. The link you provide is NOT a statement by the American Statistical Association. It is an editorial by several authors.

There is considerable debate over this issue. There are problems with p-values. However, as the authors state themselves, much of the problem is over people’s mindsets about how to use p-values and their incorrect interpretations about what statistical significance does and does not mean.

If you were to read my website more thoroughly, you’d be aware that I share many of their concerns and I address them in multiple posts. One of the authors’ key points is the need to be thoughtful and conduct thoughtful research and analysis. I emphasize this aspect in multiple posts on this topic. I’ll ask you to read the following three because they all address some of the authors’ concerns and suggestions. But you might run across others to read as well.

Five Tips for Using P-values to Avoid Being Misled How to Interpret P-values Correctly P-values and the Reproducibility of Experimental Results

September 24, 2020 at 11:52 pm

HI Jim, i just want you to know that you made explanation for Statistics so simple! I should say lesser and fewer words that reduce the complexity. All the best! 🙂

September 25, 2020 at 1:03 am

Thanks, Rene! Your kind words mean a lot to me! I’m so glad it has been helpful!

September 23, 2020 at 2:21 am

Honestly, I never understood stats during my entire M.Ed course and was another nightmare for me. But how easily you have explained each concept, I have understood stats way beyond my imagination. Thank you so much for helping ignorant research scholars like us. Looking forward to get hardcopy of your book. Kindly tell is it available through flipkart?

September 24, 2020 at 11:14 pm

I’m so happy to hear that my website has been helpful!

I checked on flipkart and it appears like my books are not available there. I’m never exactly sure where they’re available due to the vagaries of different distribution channels. They are available on Amazon in India.

Introduction to Statistics: An Intuitive Guide (Amazon IN) Hypothesis Testing: An Intuitive Guide (Amazon IN)

July 26, 2020 at 11:57 am

Dear Jim I am a teacher from India . I don’t have any background in statistics, and still I should tell that in a single read I can follow your explanations . I take my entire biostatistics class for botany graduates with your explanations. Thanks a lot. May I know how I can avail your books in India

July 28, 2020 at 12:31 am

Right now my books are only available as ebooks from my website. However, soon I’ll have some exciting news about other ways to obtain it. Stay tuned! I’ll announce it on my email list. If you’re not already on it, you can sign up using the form that is in the right margin of my website.

June 22, 2020 at 2:02 pm

Also can you please let me if this book covers topics like EDA and principal component analysis?

June 22, 2020 at 2:07 pm

This book doesn’t cover principal components analysis. Although, I wouldn’t really classify that as a hypothesis test. In the future, I might write a multivariate analysis book that would cover this and others. But, that’s well down the road.

My Introduction to Statistics covers EDA. That’s the largely graphical look at your data that you often do prior to hypothesis testing. The Introduction book perfectly leads right into the Hypothesis Testing book.

June 22, 2020 at 1:45 pm

Thanks for the detailed explanation. It does clear my doubts. I saw that your book related to hypothesis testing has the topics that I am studying currently. I am looking forward to purchasing it.

Regards, Take Care

June 19, 2020 at 1:03 pm

For this particular article I did not understand a couple of statements and it would great if you could help: 1)”If sample error causes the observed difference, the next time someone performs the same experiment the results might be different.” 2)”If the difference does not exist at the population level, you won’t obtain the benefits that you expect based on the sample statistics.”

I discovered your articles by chance and now I keep coming back to read & understand statistical concepts. These articles are very informative & easy to digest. Thanks for the simplifying things.

June 20, 2020 at 9:53 pm

I’m so happy to hear that you’ve found my website to be helpful!

To answer your questions, keep in mind that a central tenant of inferential statistics is that the random sample that a study drew was only one of an infinite number of possible it could’ve drawn. Each random sample produces different results. Most results will cluster around the population value assuming they used good methodology. However, random sampling error always exists and makes it so that population estimates from a sample almost never exactly equal the correct population value.

So, imagine that we’re studying a medication and comparing the treatment and control groups. Suppose that the medicine is truly not effect and that the population difference between the treatment and control group is zero (i.e., no difference.) Despite the true difference being zero, most sample estimates will show some degree of either a positive or negative effect thanks to random sampling error. So, just because a study has an observed difference does not mean that a difference exists at the population level. So, on to your questions:

1. If the observed difference is just random error, then it makes sense that if you collected another random sample, the difference could change. It could change from negative to positive, positive to negative, more extreme, less extreme, etc. However, if the difference exists at the population level, most random samples drawn from the population will reflect that difference. If the medicine has an effect, most random samples will reflect that fact and not bounce around on both sides of zero as much.

2. This is closely related to the previous answer. If there is no difference at the population level, but say you approve the medicine because of the observed effects in a sample. Even though your random sample showed an effect (which was really random error), that effect doesn’t exist. So, when you start using it on a larger scale, people won’t benefit from the medicine. That’s why it’s important to separate out what is easily explained by random error versus what is not easily explained by it.

I think reading my post about how hypothesis tests work will help clarify this process. Also, in about 24 hours (as I write this), I’ll be releasing my new ebook about Hypothesis Testing!

May 29, 2020 at 5:23 am

Hi Jim, I really enjoy your blog. Can you please link me on your blog where you discuss about Subgroup analysis and how it is done? I need to use non parametric and parametric statistical methods for my work and also do subgroup analysis in order to identify potential groups of patients that may benefit more from using a treatment than other groups.

May 29, 2020 at 2:12 pm

Hi, I don’t have a specific article about subgroup analysis. However, subgroup analysis is just the dividing up of a larger sample into subgroups and then analyzing those subgroups separately. You can use the various analyses I write about on the subgroups.

Alternatively, you can include the subgroups in regression analysis as an indicator variable and include that variable as a main effect and an interaction effect to see how the relationships vary by subgroup without needing to subdivide your data. I write about that approach in my article about comparing regression lines . This approach is my preferred approach when possible.

April 19, 2020 at 7:58 am

sir is confidence interval is a part of estimation?

April 17, 2020 at 3:36 pm

Sir can u plz briefly explain alternatives of hypothesis testing? I m unable to find the answer

April 18, 2020 at 1:22 am

Assuming you want to draw conclusions about populations by using samples (i.e., inferential statistics ), you can use confidence intervals and bootstrap methods as alternatives to the traditional hypothesis testing methods.

March 9, 2020 at 10:01 pm

Hi JIm, could you please help with activities that can best teach concepts of hypothesis testing through simulation, Also, do you have any question set that would enhance students intuition why learning hypothesis testing as a topic in introductory statistics. Thanks.

March 5, 2020 at 3:48 pm

Hi Jim, I’m studying multiple hypothesis testing & was wondering if you had any material that would be relevant. I’m more trying to understand how testing multiple samples simultaneously affects your results & more on the Bonferroni Correction

March 5, 2020 at 4:05 pm

I write about multiple comparisons (aka post hoc tests) in the ANOVA context . I don’t talk about Bonferroni Corrections specifically but I cover related types of corrections. I’m not sure if that exactly addresses what you want to know but is probably the closest I have already written. I hope it helps!

January 14, 2020 at 9:03 pm

Thank you! Have a great day/evening.

January 13, 2020 at 7:10 pm

Any help would be greatly appreciated. What is the difference between The Hypothesis Test and The Statistical Test of Hypothesis?

January 14, 2020 at 11:02 am

They sound like the same thing to me. Unless this is specialized terminology for a particular field or the author was intending something specific, I’d guess they’re one and the same.

April 1, 2019 at 10:00 am

so these are the only two forms of Hypothesis used in statistical testing?

April 1, 2019 at 10:02 am

Are you referring to the null and alternative hypothesis? If so, yes, that’s those are the standard hypotheses in a statistical hypothesis test.

April 1, 2019 at 9:57 am

year very insightful post, thanks for the write up

October 27, 2018 at 11:09 pm

hi there, am upcoming statistician, out of all blogs that i have read, i have found this one more useful as long as my problem is concerned. thanks so much

October 27, 2018 at 11:14 pm

Hi Stano, you’re very welcome! Thanks for your kind words. They mean a lot! I’m happy to hear that my posts were able to help you. I’m sure you will be a fantastic statistician. Best of luck with your studies!

October 26, 2018 at 11:39 am

Dear Jim, thank you very much for your explanations! I have a question. Can I use t-test to compare two samples in case each of them have right bias?

October 26, 2018 at 12:00 pm

Hi Tetyana,

You’re very welcome!

The term “right bias” is not a standard term. Do you by chance mean right skewed distributions? In other words, if you plot the distribution for each group on a histogram they have longer right tails? These are not the symmetrical bell-shape curves of the normal distribution.

If that’s the case, yes you can as long as you exceed a specific sample size within each group. I include a table that contains these sample size requirements in my post about nonparametric vs parametric analyses .

Bias in statistics refers to cases where an estimate of a value is systematically higher or lower than the true value. If this is the case, you might be able to use t-tests, but you’d need to be sure to understand the nature of the bias so you would understand what the results are really indicating.

I hope this helps!

April 2, 2018 at 7:28 am

Simple and upto the point 👍 Thank you so much.

April 2, 2018 at 11:11 am

Hi Kalpana, thanks! And I’m glad it was helpful!

March 26, 2018 at 8:41 am

Am I correct if I say: Alpha – Probability of wrongly rejection of null hypothesis P-value – Probability of wrongly acceptance of null hypothesis

March 28, 2018 at 3:14 pm

You’re correct about alpha. Alpha is the probability of rejecting the null hypothesis when the null is true.

Unfortunately, your definition of the p-value is a bit off. The p-value has a fairly convoluted definition. It is the probability of obtaining the effect observed in a sample, or more extreme, if the null hypothesis is true. The p-value does NOT indicate the probability that either the null or alternative is true or false. Although, those are very common misinterpretations. To learn more, read my post about how to interpret p-values correctly .

March 2, 2018 at 6:10 pm

I recently started reading your blog and it is very helpful to understand each concept of statistical tests in easy way with some good examples. Also, I recommend to other people go through all these blogs which you posted. Specially for those people who have not statistical background and they are facing to many problems while studying statistical analysis.

Thank you for your such good blogs.

March 3, 2018 at 10:12 pm

Hi Amit, I’m so glad that my blog posts have been helpful for you! It means a lot to me that you took the time to write such a nice comment! Also, thanks for recommending by blog to others! I try really hard to write posts about statistics that are easy to understand.

January 17, 2018 at 7:03 am

I recently started reading your blog and I find it very interesting. I am learning statistics by my own, and I generally do many google search to understand the concepts. So this blog is quite helpful for me, as it have most of the content which I am looking for.

January 17, 2018 at 3:56 pm

Hi Shashank, thank you! And, I’m very glad to hear that my blog is helpful!

January 2, 2018 at 2:28 pm

thank u very much sir.

January 2, 2018 at 2:36 pm

You’re very welcome, Hiral!

November 21, 2017 at 12:43 pm

Thank u so much sir….your posts always helps me to be a #statistician

November 21, 2017 at 2:40 pm

Hi Sachin, you’re very welcome! I’m happy that you find my posts to be helpful!

November 19, 2017 at 8:22 pm

great post as usual, but it would be nice to see an example.

November 19, 2017 at 8:27 pm

Thank you! At the end of this post, I have links to four other posts that show examples of hypothesis tests in action. You’ll find what you’re looking for in those posts!

Comments and Questions Cancel reply

7.1 The Central Limit Theorem for Sample Means (Averages)

Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose:

• μ X = the mean of X
• σ X = the standard deviation of X

If you draw random samples of size n , then as n increases, the random variable x ¯ x ¯ which consists of sample means, tends to be normally distributed and

The central limit theorem for sample means says that if you repeatedly draw samples of a given size (such as repeatedly rolling ten dice) and calculate their means, those means tend to follow a normal distribution (the sampling distribution). As sample sizes increase, the distribution of means more closely follows the normal distribution. The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. Standard deviation is the square root of variance, so the standard deviation of the sampling distribution is the standard deviation of the original distribution divided by the square root of n . The variable n is the number of values that are averaged together, not the number of times the experiment is done.

To put it more formally, if you draw random samples of size n , the distribution of the random variable X ¯ X ¯ , which consists of sample means, is called the sampling distribution of the mean . The sampling distribution of the mean approaches a normal distribution as n , the sample size , increases.

The random variable X ¯ X ¯ has a different z -score associated with it from that of the random variable X . The mean x ¯ x ¯ is the value of X ¯ X ¯ in one sample.

μ X is the average of both X and X ¯ X ¯ .

σ x = σ X n = σ x = σ X n = standard deviation of X ¯ X ¯ and is called the standard error of the mean.

Using the TI-83, 83+, 84, 84+ Calculator

To find probabilities for means on the calculator, follow these steps.

2nd DISTR 2:normalcdf

n o r m a l c d f ( l o w e r   v a l u e   o f   t h e   a r e a ,   u p p e r   v a l u e   o f   t h e   a r e a ,   m e a n ,   s t a n d a r d   d e v i a t i o n s a m p l e   s i z e ) n o r m a l c d f ( l o w e r   v a l u e   o f   t h e   a r e a ,   u p p e r   v a l u e   o f   t h e   a r e a ,   m e a n ,   s t a n d a r d   d e v i a t i o n s a m p l e   s i z e )

• mean is the mean of the original distribution
• standard deviation is the standard deviation of the original distribution
• sample size = n

Example 7.1

An unknown distribution has a mean of 90 and a standard deviation of 15. Samples of size n = 25 are drawn randomly from the population.

a. Find the probability that the sample mean is between 85 and 92.

a. Let X = one value from the original unknown population. The probability question asks you to find a probability for the sample mean .

Let x ¯ x ¯ = the mean of a sample of size 25. Since μ X = 90, σ X = 15, and n = 25,

x ¯ x ¯ ~ N ( 90 ,  15 25 ) ( 90 ,  15 25 ) .

Find P (85 < x ¯ x ¯ < 92). Draw a graph.

P (85 < x ¯ x ¯ < 92) = 0.6997

The probability that the sample mean is between 85 and 92 is 0.6997.

normalcdf (lower value, upper value, mean, standard error of the mean)

The parameter list is abbreviated (lower value, upper value, μ , σ n σ n )

normalcdf (85,92,90, 15 25 15 25 ) = 0.6997

b. Find the value that is two standard deviations above the expected value, 90, of the sample mean.

b. To find the value that is two standard deviations above the expected value 90, use the formula:

value = μ x + (#ofTSDEVs) ( σ x n ) ( σ x n )

value = 90 + 2 ( 15 25 ) ( 15 25 ) = 96

The value that is two standard deviations above the expected value is 96.

The standard error of the mean is σ x n σ x n = 15 25 15 25 = 3. Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size n .

An unknown distribution has a mean of 45 and a standard deviation of eight. Samples of size n = 30 are drawn randomly from the population. Find the probability that the sample mean is between 42 and 50.

Example 7.2

The length of time, in hours, it takes an "over 40" group of people to play one soccer match is normally distributed with a mean of two hours and a standard deviation of 0.5 hours . A sample of size n = 50 is drawn randomly from the population. Find the probability that the sample mean is between 1.8 hours and 2.3 hours.

Let X = the time, in hours, it takes to play one soccer match.

The probability question asks you to find a probability for the sample mean time, in hours , it takes to play one soccer match.

Let x ¯ x ¯ = the mean time, in hours, it takes to play one soccer match.

If μ X = _________, σ X = __________, and n = ___________, then X ¯ X ¯ ~ N (______, ______) by the central limit theorem for means .

μ X = 2, σ X = 0.5, n = 50, and X ~ N ( 2,  0.5 50 ) ( 2,  0.5 50 )

Find P (1.8 < x ¯ x ¯ < 2.3). Draw a graph.

P (1.8 < x ¯ x ¯ < 2.3) = 0.9977

normalcdf ( 1. 8,2 .3,2, .5 50 ) ( 1. 8,2 .3,2, .5 50 ) = 0.9977

The probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977.

The length of time taken on the SAT for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of n = 60 is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours.

To find percentiles for means on the calculator, follow these steps.

2 nd DIStR 3:invNorm

k = invNorm ( area to the left of  k , mean,   s t a n d a r d   d e v i a t i o n s a m p l e   s i z e ) ( area to the left of  k , mean,   s t a n d a r d   d e v i a t i o n s a m p l e   s i z e )

• k = the k th percentile

Example 7.3

In a recent study, it was reported that the mean age of iPad users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size n = 100.

• What are the mean and standard deviation for the sample mean ages of iPad users?
• What does the distribution look like?
• Find the probability that the sample mean age is more than 30 years (the reported mean age of iPad users in this particular study).
• Find the 95 th percentile for the sample mean age (to one decimal place).
• Since the sample mean tends to target the population mean, we have μ χ = μ = 34. The sample standard deviation is given by σ χ = σ n σ n = 15 100 15 100 = 15 10 15 10 = 1.5
• The central limit theorem states that for large sample sizes( n ), the sampling distribution will be approximately normal.
• The probability that the sample mean age is more than 30 is given by P ( X ¯   >   30 ) P ( X ¯   >   30 ) = normalcdf (30,E99,34,1.5) = 0.9962
• Let k = the 95 th percentile. k = invNorm ( 0. 95,34, 15 100 ) ( 0. 95,34, 15 100 ) = 36.5

In an article on Flurry Blog, a gaming marketing gap for men between the ages of 30 and 40 is identified. You are researching a startup game targeted at the 35-year-old demographic. Your idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. Based on the article’s data, industry research shows that the average strategy player is 28 years old with a standard deviation of 4.8 years. You take a sample of 100 randomly selected gamers. If your target market is 29- to 35-year-olds, should you continue with your development strategy?

Example 7.4

The mean number of minutes for app engagement by an iPad user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60.

• What are the mean and standard deviation for the sample mean number of minutes for app engagement by an iPad user?
• What is the standard error of the mean?
• Find the 90 th percentile for the sample mean time of minutes for app engagement by an iPad user. Interpret this value in a complete sentence.
• Find the probability that the sample mean is between eight minutes and 8.5 minutes.
• μ x ¯ = μ = 8.2   σ x ¯ = σ n = 1 60 = 0.13 μ x ¯ = μ = 8.2   σ x ¯ = σ n = 1 60 = 0.13
• This allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60.
• Let k = the 90 th percentile k = invNorm ( 0. 90,8 .2, 1 60 ) ( 0. 90,8 .2, 1 60 ) = 8.37. This values indicates that 90 percent of the average app engagement time for iPad users is less than 8.37 minutes.
• P (8 < x ¯ x ¯ < 8.5) = normalcdf ( 8,8 .5,8 .2, 1 60 ) ( 8,8 .5,8 .2, 1 60 ) = 0.9293

Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are n = 34, x ¯ x ¯ = 16.01 ounces. If the cans are filled so that μ = 16.00 ounces (as labeled) and σ = 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces?

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4.1 - sampling distribution of the sample mean.

In the following example, we illustrate the sampling distribution for the sample mean for a very small population. The sampling method is done without replacement.

Sample Means with a Small Population: Pumpkin Weights

In this example, the population is the weight of six pumpkins (in pounds) displayed in a carnival "guess the weight" game booth. You are asked to guess the average weight of the six pumpkins by taking a random sample without replacement from the population.

Since we know the weights from the population, we can find the population mean.

\(\mu=\dfrac{19+14+15+9+10+17}{6}=14\) pounds

To demonstrate the sampling distribution, let’s start with obtaining all of the possible samples of size \(n=2\) from the populations, sampling without replacement. The table below shows all the possible samples, the weights for the chosen pumpkins, the sample mean and the probability of obtaining each sample. Since we are drawing at random, each sample will have the same probability of being chosen.

We can combine all of the values and create a table of the possible values and their respective probabilities.

The table is the probability table for the sample mean and it is the sampling distribution of the sample mean weights of the pumpkins when the sample size is 2. It is also worth noting that the sum of all the probabilities equals 1. It might be helpful to graph these values.

One can see that the chance that the sample mean is exactly the population mean is only 1 in 15, very small. (In some other examples, it may happen that the sample mean can never be the same value as the population mean.) When using the sample mean to estimate the population mean, some possible error will be involved since the sample mean is random.

Now that we have the sampling distribution of the sample mean, we can calculate the mean of all the sample means. In other words, we can find the mean (or expected value) of all the possible \(\bar{x}\)’s.

The mean of the sample means is

\(\mu_\bar{x}=\sum \bar{x}_{i}f(\bar{x}_i)=9.5\left(\frac{1}{15}\right)+11.5\left(\frac{1}{15}\right)+12\left(\frac{2}{15}\right)\\+12.5\left(\frac{1}{15}\right)+13\left(\frac{1}{15}\right)+13.5\left(\frac{1}{15}\right)+14\left(\frac{1}{15}\right)\\+14.5\left(\frac{2}{15}\right)+15.5\left(\frac{1}{15}\right)+16\left(\frac{1}{15}\right)+16.5\left(\frac{1}{15}\right)\\+17\left(\frac{1}{15}\right)+18\left(\frac{1}{15}\right)=14\)

Even though each sample may give you an answer involving some error, the expected value is right at the target: exactly the population mean. In other words, if one does the experiment over and over again, the overall average of the sample mean is exactly the population mean.

Now, let's do the same thing as above but with sample size \(n=5\)

The sampling distribution is:

The mean of the sample means is...

\(\mu=(\dfrac{1}{6})(13+13.4+13.8+14.0+14.8+15.0)=14\) pounds

The following dot plots show the distribution of the sample means corresponding to sample sizes of \(n=2\) and of \(n=5\).

Again, we see that using the sample mean to estimate population mean involves sampling error. However, the error with a sample of size \(n=5\) is on the average smaller than with a sample of size \(n= 2\).

Sampling Error and Size

Sample size and sampling error: As the dotplots above show, the possible sample means cluster more closely around the population mean as the sample size increases. Thus, the possible sampling error decreases as sample size increases.

What happens when the population is not small, as in the pumpkin example?

Sample Means with Large Samples: Exam Example

An instructor of an introduction to statistics course has 200 students. The scores out of 100 points are shown in the histogram.

The population mean is \(μ=71.18\) and the population standard deviation is \(σ=10.73\)

Let's demonstrate the sampling distribution of the sample means using the StatKey website . The first video will demonstrate the sampling distribution of the sample mean when n = 10 for the exam scores data. The second video will show the same data but with samples of n = 30.

You should start to see some patterns. The mean of the sampling distribution is very close to the population mean. The standard deviation of the sampling distribution is smaller than the standard deviation of the population.

In the examples so far, we were given the population and sampled from that population.

What happens when we do not have the population to sample from? What happens when all that we are given is the sample? Fortunately, we can use some theory to help us. The mathematical details of the theory are beyond the scope of this course but the results are presented in this lesson.

In the next two sections, we will discuss the sampling distribution of the sample mean when the population is Normally distributed and when it is not.

Sample Mean: Symbol (X Bar), Definition, Standard Error

Contents (click to go to the section):

• Sample Mean Symbol

How to Find the Sample Mean

• Variance of the sampling distribution of the sample mean

Calculate Standard Error for the Sample Mean

Watch the video for an example of how to find the sample mean:

Can’t see the video? Click here to watch it on YouTube.

Sample Mean Symbol and Definition

The sample mean symbol is x̄, pronounced “x bar”.

The sample mean is useful because it allows you to estimate what the whole population is doing, without surveying everyone. Let’s say your sample mean for the food example was $2400 per year. The odds are, you would get a very similar figure if you surveyed all 300 million people. So the sample mean is a way of saving a lot of time and money.

The sample mean formula is:

x̄ = ( Σ x i ) / n

If that looks complicated, it’s simpler than you think (although check out our tutoring page if you need help!). Remember the formula to find an “ average ” in basic math? It’s the exact same thing, only the notation (i.e. the symbols) are just different. Let’s break it down into parts:

• x̄ just stands for the “sample mean”
• Σ is summation notation , which means “add up”
• x i “all of the x-values”
• n means “the number of items in the sample”

Now it’s just a matter of plugging in numbers that you’re given and solving using arithmetic (there’s no algebra required—you can basically plug this in to any calculator).

You might see the following alternate sample mean formula : x̄ = 1/ n * ( Σ x i ) The set up is slightly different, but algebraically it’s the same formula (if you simplify the formula 1/n * X, you get 1/X).

Back to Top

Finding the sample mean is no different from finding the average of a set of numbers. In statistics you’ll come across slightly different notation than you’re probably used to, but the math is exactly the same.

The formula to find the sample mean is: = ( Σ x i ) / n.

All that formula is saying is add up all of the numbers in your data set ( Σ means “add up” and x i means “all the numbers in the data set). This article tells you how to find the sample mean by hand (this is also one of the AP Statistics formulas ). However, if you’re finding the sample mean, you’re probably going to be finding other descriptive statistics, like the sample variance or the interquartile range so you may want to consider finding the sample mean in Excel or other technology. Why? Although the calculation for the mean is fairly simple, if you use Excel then you only have to enter the numbers once. After that, you can use the numbers to find any statistic: not just the sample mean.

How to Find the Sample Mean: Steps

Sample Question: Find the sample mean for the following set of numbers: 12, 13, 14, 16, 17, 40, 43, 55, 56, 67, 78, 78, 79, 80, 81, 90, 99, 101, 102, 304, 306, 400, 401, 403, 404, 405.

Step 1: Add up all of the numbers : 12 + 13 + 14 + 16 + 17 + 40 + 43 + 55 + 56 + 67 + 78 + 78 + 79 + 80 + 81 + 90 + 99 + 101 + 102 + 304 + 306 + 400 + 401 + 403 + 404 + 405 = 3744 .

Step 2: Count the numbers of items in your data set . In this particular data set there are 26 items.

Step 3: Divide the number you found in Step 1 by the number you found in Step 2. 3744/26 = 144.

That’s it!

Tip: If you have to show working out on a test, just place the two numbers into the formula. Step 1 gives you the σ and Step 2 gives you n: x = ( Σ x i ) / n = 3744/26 = 144

Variance of the Sampling Distribution of the Sample Mean

This section covers the variance of the sampling distribution of the mean. If you aren’t familiar with the central limit theorem , you may want to read the previous article: The Mean of the Sampling Distribution of the Mean .

Watch the video or read on below:

The sampling distribution of the sample mean is a probability distribution of all the sample means. Let’s say you had 1,000 people, and you sampled 5 people at a time and calculated their average height. If you kept on taking samples (i.e. you repeated the sampling a thousand times), eventually the mean of all of your sample means will:

• Equal the population mean , μ
• Look like a normal distribution curve.

The variance of this probability distribution gives you an idea of how spread out your data is around the mean . The larger the sample size, the more closely the sample mean will represent the population mean . In other words, as N grows larger, the variance becomes smaller. Ideally, when the sample mean matches the population mean, the variance will equal zero.

The formula to find the variance of the sampling distribution of the mean is: σ 2 M = σ 2 / N, where: σ 2 M = variance of the sampling distribution of the sample mean. σ 2 = population variance . N = your sample size.

Sample question: If a random sample of size 19 is drawn from a population distribution with standard deviation α = 20 then what will be the variance of the sampling distribution of the sample mean?

Step 1: Figure out the population variance . Variance is the standard deviation squared, so: σ 2 = 20 2 = 400.

Step 2: Divide the variance by the number of items in the sample. This sample has 19 items, so: 400 / 19 = 21.05.

How to Calculate Standard Error for the Sample Mean: Overview

The standard error of the mean of a sample is equal to the standard deviation for the sample . The difference between standard error and standard deviation is that with standard deviations you use population data (i.e. parameters ) and with standard errors you use data from your sample. You can calculate standard error for the sample mean using the formula:

SE = s / √(n)

SE = standard error, s = the standard deviation for your sample and n is the number of items in your sample.

Calculate Standard Error for the Sample Mean: Steps

Example: Find the standard error for the following heights (in cm): Jim (170.5), John (161), Jack (160), Freda (170), Tai (150.5).

Step 1: Find the mean (the average ) of the data set: (170.5 + 161 + 160 + 170 + 150.5) / 5 = 162.4.

Step 2: Calculate the deviation from the mean by subtracting each value from the mean you found in Step 1. 170.5 – 162.4 = -8.1 161 – 162.4 = 1.4 160 – 162.4 = 2.4 170 – 162.4 = -7.6 150.5 – 162.4 = 11.9

Step 3: Square the numbers you calculated in Step 2:

-8.1 * -8.1 = 65.61 1.4 * 1.4 = 1.96 2.4 * 2.4 = 5.76 -7.6 * -7.6 = 57.76 11.9 * 11.9 = 141.61

Step 4: Add the values you calculated in Step 3: 65.61 + 1.96 + 5.76 + 57.76 + 141.61 = 272.7

Step 5: Divide the number you found in Step 4 by your sample size – 1 . There are five items in the sample, so n-1 = 4: 272.7 / 4 = 68.175.

Step 6: Take the square root of the number you found in Step 5. This is your standard deviation. √(68.175) = 8.257

Step 6: Divide the number you calculated in Step 6 by the square root of the sample size (in this sample problem, the sample size is 5): 8.257 / √(5) = 8.257 / 2.236 = 3.693

That’s how to calculate the standard error for the sample mean!

Tip: If you’re asked to find the “standard error” for a sample, in most cases you’re finding the sample error for the mean using the formula SE = s/√n. There are different types of standard error though (i.e. for proportions), so you may want to make sure you’re calculating the right statistic.

Evans, M.; Hastings, N.; and Peacock, B. Statistical Distributions, 3rd ed. New York: Wiley, p. 16, 2000. Kenney, J. F. and Keeping, E. S. “Averages,” “Relation Between Mean, Median, and Mode,” and “Relative Merits of Mean, Median, and Mode.” §3.1 and §4.8-4.9 in Mathematics of Statistics, Pt. 1, 3rd ed. Princeton, NJ: Van Nostrand, pp. 32 and 52-54, 1962.